Given the Dirichlet beta function,

$$\beta(k) = \sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^k}$$

(The cases k = 2 is Catalan's constant.) It seems,

$$\sum_{k=2}^\infty\Big[1-\beta(k) \Big] = \frac{1}{4}\big(\pi+\log(4)-4\big)=0.131971\dots$$

or, in general, for some constant p > 0,

$$\sum_{k=2}^\infty\left[1-\sum_{n=0}^\infty\frac{(-1)^n}{(pn+1)^k} \right] = \sum_{m=1}^\infty\frac{1}{2p^2m^2+3pm+1}$$

Anyone knows how to prove the general proposed equality? (This is similar to the question on the zeta sum here.)


Here is a way to derive a slightly different looking result:

Notice that $$\sum_{k=2}^{\infty}\left[1-\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(pn+1)^{k}}\right]=\sum_{k=2}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(pn+1)^{k}}$$

$$=\sum_{n=1}^{\infty}(-1)^{n-1}\sum_{k=2}^{\infty}\frac{1}{(pn+1)^{k}}=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(pn+1)^{2}}\sum_{k=0}^{\infty}\frac{1}{(pn+1)^{k}}.$$ Now, since $$\sum_{k=0}^{\infty}\frac{1}{(pn+1)^{k}}=\frac{1}{1-\frac{1}{pn+1}}=\frac{pn+1}{pn},$$ our series is


Plugging in the case $p=2$ seems to agree with your first identity.

Remark: Using partial fractions, we can go a bit further. Notice that $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{pn(pn+1)}=\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{1}{pn}-\frac{1}{pn+1}\right)=\frac{\log 2}{p}-\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{pn+1} $$

Suppose $p$ is an integer, and let $\zeta_{p}$ be a $p^{th}$ root of unity. Then consider $$\frac{\log\left(1+z\right)}{z}+\frac{\log\left(1+\zeta_{p}z\right)}{\zeta_{p}z}+\cdots+\frac{\log\left(1+\zeta_{p}^{p-1}z\right)}{\zeta_{p}^{p-1}z}=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{n}z^{n-1}\sum_{k=0}^{p-1}\zeta_{p}^{k(n-1)} $$

$$=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{pn+1}z^{pn}.$$ Letting $z=1,$ we have the identity $$\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{pn+1}=\sum_{k=0}^{p-1}\frac{\log\left(1+\zeta_{p}^{k}\right)}{\zeta_{p}^{k}z},$$ so our original series is $$\frac{1}{p}\log2+\sum_{k=0}^{p-1}\frac{\log\left(1+\zeta_{p}^{k}\right)}{\zeta_{p}^{k}z}.$$

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    $\begingroup$ That was fast. :-) WolframAlpha says my non-alternating series and your alternating one are in fact equal. $\endgroup$ – Tito Piezas III Jun 24 '12 at 16:44
  • $\begingroup$ @Tito: Ahhh, good point. $\endgroup$ – Eric Naslund Jun 24 '12 at 17:11
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    $\begingroup$ At the end, you said, "Letting $z=1$," but there's still a $z$ there. Is that a typo? $\endgroup$ – Akiva Weinberger Aug 29 '14 at 12:28

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